I was trying to prove that $\lim_{x\to3}(x^2)=9$ and read the answer sheet, feeling confused. I could not understand the explanation below.
If $0 < |x - 3| < \delta$ then $|x + 3||x - 3| < \epsilon.$
We can find a positive constant $C$ such that $|x + 3| < C,$ then $|x + 3||x - 3| < C|x - 3|.$
Then we can make $C|x - 3| < \epsilon$ by taking $|x - 3| < \epsilon / C = \delta.$
Here is my question: Why does $C|x - 3|$ become less than $\epsilon?$ And why do we divide $\epsilon$ by the constant $C?$ I just don't understand why we are manipulating the equation this way here. Here's the next part of the explanation:
We can find such a number $C$ if we restrict $x$ to lie in some interval centered at $3$. In fact, since we are interested only in values of $x$ that are close to $3$, it is reasonable to assume that $x$ is within a distance $1$ from $3$, that is, $|x - 3| < 1.$
Here's my second question: how do we know that $x$ is within a distance of $1$ from $3?$ Couldn't this distance be any other number?
Then $2 < x < 4,$ so $5 < x + 3 < 7.$ Thus we have $|x + 3| < 7,$ and so $C = 7$ is a suitable choice for the constant.
The last question: Why does $|x + 3| < 7$ mean that we can choose $C$ as $7?$
Later on, the explanation mentions two restrictions on $|x - 3|$ and makes $\delta$ the smaller of the two numbers $1$ and $\epsilon / 7,$ but I think I might be able to figure that out on my own if I can understand the explanation above. Most of all I don't even understand the purpose of the constant $C,$ and why it is used to divide $\epsilon.$ Anyone please explain this concept in an easy way!
- The explanation is from Single Variable Calculus by James Stewart.
First Question: We set $\delta=\epsilon/C$ Hence $|x-3|<\epsilon/C$ and so $C|x-3|<\epsilon$. Now notice that if we have bound $|x+3|<C$ then $|x+3||x-3|<C|x-3|<\epsilon$. Hence we have shown the limit is $9$ from definition. So if we can show that $|x+3|<C$ we can prove the limit by simply setting $\delta=\epsilon/C$. That is why we try to find $C$
Second Question: It could be a different number, virtually any other positive real number, $1$ was chosen for simplicity. Let's pick $2$, then we get $|x-3|<2$ and so $1<x<5$ and so $4<x+3<8$ and now our C is $8$. But yes, picking 1 just makes things simpler.
Last question: We did not "choose" C. We have proven that C is at most 7. The whole point of this operation is that changing $\delta$ in an obvious way manipulates how big $|x-3|$ is, but it is not clear how it changes $|x+3|$ so we bound $|x+3|$ by C so that we do not have to worry about it.